What is the Ratio of Internal to External Diameter for Equal Strength in Shafts?

Click For Summary

Homework Help Overview

The discussion revolves around determining the ratio of internal to external diameter for a hollow shaft made of stainless steel, intended to replace a solid mild steel shaft, while maintaining equal strength. The problem involves concepts of torsion and shear stress in materials, specifically comparing mild steel and stainless steel under torsional loads.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between maximum shear stress and torsional strength, attempting to set up equations based on these principles. There are questions about the correctness of algebraic manipulations and the implications of negative signs in calculations. Some participants express confusion regarding how to derive a ratio from their equations.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's algebraic approaches. There is a recognition of differing interpretations of the calculations, and some participants are questioning the validity of the results obtained. No consensus has been reached regarding the correct ratio or the steps to derive it.

Contextual Notes

Participants note that the densities of mild steel and stainless steel are assumed to be equal for the purpose of this problem. There is also mention of the maximum shear stress values for both materials, which are critical to the calculations being discussed.

MMCS
Messages
151
Reaction score
0
A mild steel solid shaft is to be replaced by a stainless steel hollow shaft of the same
outside diameter. Calculate the ratio of internal diameter to external diameter of the
hollow shaft for equal strength. Also find the percentage saving in weight if density of
mild steel is the same as stainless steel.Take maximum shear stress for Mild Steel as
34 MPa and that for Stainless Steel as 57 MPa

I assume that for the strength to be the same the max torsion should be the same so i use the max allowed shear stress in this equationrearranged for torsion and equate them both

(34MPa * (pi*R^4)/2)/R = (57MPa * (pi*(R-Ri)^4)/2)/R - Ri

RI = inside radius

Is this the correct step? how would i get a ratio from this?
 
Physics news on Phys.org
MMCS said:
A mild steel solid shaft is to be replaced by a stainless steel hollow shaft of the same
outside diameter. Calculate the ratio of internal diameter to external diameter of the
hollow shaft for equal strength. Also find the percentage saving in weight if density of
mild steel is the same as stainless steel.Take maximum shear stress for Mild Steel as
34 MPa and that for Stainless Steel as 57 MPa

I assume that for the strength to be the same the max torsion should be the same so i use the max allowed shear stress in this equationrearranged for torsion and equate them both

(34MPa * (pi*R^4)/2)/R = (57MPa * (pi*(R-Ri)^4)/2)/R - Ri

RI = inside radius

Is this the correct step? how would i get a ratio from this?
(R^4 - Ri^4) is not the same as (R - Ri)^4.
And in both cases, max shear stress occurs at the same distance fron the centroid of the circular cross section.
 
Ok so

(34MPa * (pi*R^4)/2)/R = (57MPa * (pi/2*R^4)-(pi/2*Ri^4))/R

to simplify and multiply out pi terms i get

(-23*pi/2*r^4)/r = (57*pi/2*ri^4)/r

-36.128*r^4 = 89.535ri^4

so as a ratio i get

36.128/89.535 = 0.404

but i have the answer to be 0.8
 
You've got negative signs appearing and disappearing at random.

You should review basic algebra.
 
SteamKing said:
You've got negative signs appearing and disappearing at random.

You should review basic algebra.

That is irrelevant. -36.128/89.535 produces the same ratio as 36.128/89.535.
 
And yet you have not solved this problem.
 
SteamKing said:
And yet you have not solved this problem.

Another irrelevant response. I do not know how to solve the problem otherwise i wouldn't have posted the question, however, Your input had nothing to do with getting closer to the solution.
 
MMCS said:
Ok so

(34MPa * (pi*R^4)/2)/R = (57MPa * (pi/2*R^4)-(pi/2*Ri^4))/R

to simplify and multiply out pi terms i get

(-23*pi/2*r^4)/r = (57*pi/2*ri^4)/r

-36.128*r^4 = 89.535ri^4

so as a ratio i get

36.128/89.535 = 0.404
the plus and minus signs heretofore not withstanding, you have calculated that (ri^4/r^4) = (ri/r)^4 = 0.404. So to calculate (ri/r), you need to take the 4th root of that number.
but i have the answer to be 0.8
which is correct.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 17 ·
Replies
17
Views
48K
  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 24 ·
Replies
24
Views
10K
  • · Replies 3 ·
Replies
3
Views
14K
  • · Replies 5 ·
Replies
5
Views
6K
Replies
3
Views
2K
Replies
4
Views
4K